Question: Simplify the following expression and state the condition under which the simplification is valid. $q = \dfrac{6x^2 + 60x + 126}{4x^2 + 44x + 112}$
First factor out the greatest common factors in the numerator and in the denominator. $ q = \dfrac {6(x^2 + 10x + 21)} {4(x^2 + 11x + 28)} $ $ q = \dfrac{6}{4} \cdot \dfrac{x^2 + 10x + 21}{x^2 + 11x + 28} $ Simplify: $ q = \dfrac{3}{2} \cdot \dfrac{x^2 + 10x + 21}{x^2 + 11x + 28}$ Next factor the numerator and denominator. $ q = \dfrac{3}{2} \cdot \dfrac{(x + 7)(x + 3)}{(x + 7)(x + 4)}$ Assuming $x \neq -7$ , we can cancel the $x + 7$ $ q = \dfrac{3}{2} \cdot \dfrac{x + 3}{x + 4}$ Therefore: $ q = \dfrac{ 3(x + 3)}{ 2(x + 4)}$, $x \neq -7$